- #1

StephenPrivitera

- 362

- 0

F=2GMr/d

^{3}

M is the mass of the body being orbited

r is the distance from the center of the orbiting body

d is the distance between the centers of the bodies

But... Tidal force is 2F outward for r toward or away from the Sun and is -F in the plane perpendicular to this line. Why?

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So then we have tidal bulge. "Because the Earth is no longer a perfect sphere the earth exerts a torque on the moon. The moon receives a positive net torque and its orbit evolves outwards. Angular momentum is extracted from the earth rotation. As the moon gets farther away, the day lengthens."

Does this mean that because of the torque exerted on the moon, the moon gains angular momentum? And because the moon gains momentum, the earth loses it. Is angular momentum pertinent to an objects spinning or its revolution? From the above quote it sounds like the moon's orbit evolves outward due to a gain in angular momentum. But it's spinning rate doesn't increase? Since it orbits at a greater distance it must orbit more slowly, and so in fact its spinning slows down (since it is tidally locked). Is it true that the Earth exerts a torque because the force of gravity can no longer be considered to come from the center of the Earth? I'm very confused about all this.

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PR Drag

A particle that reradiates solar energy actually emits more momentum in the forward direction than behind it. This cause the object to slow down. Why? And since it slows down it spirals into the sun. Again, why? What about v=sqrt(GM/r)? As v decreases r increases.

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Yarkovski Effect

For prograde rotation, the force is positive. For retrograde the force is negative. Positive, negative?? Do you mean in the direction of motion and opposite motion? Isn't this pretty much the same as PR Drag?

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Gas Drag

When an object is larger than the mean free path of the gas it is travelling in, we can consider the gas a fluid. What is the mean free path of the gas?

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Any links to good resources on this topic would be

*greatly*appreciated.